566 10 Particle Physics – II
10.25 (a) From Problem 10.18, we have the result
∣
∣a 3 / 2
∣
∣^2 ∝190 mb
{
1
√
3∣
∣a 3 / 2 + 2 a 1 / 2∣
∣
} 2
∝70 mbDividing one by the other, and solving we finda 1 / 2 = 0. 0256 a 3 / 2. Thus
the amplitudea 1 / 2 is negligible. The resonance is therefore characterized
byI= 3 /2.
(b) The fact thatπ+pscattering cross-section has a fairly high value at
p=230 MeV/c implies thata 32 amplitude is dominant.10.26 From the analysis ofπ-N scattering (Problem 10. 18) the ratio
σ(π++p→Δ)
σ(π−+p→Δ)=
∣
∣
∣a^32∣
∣
∣
2{
√^1
3∣
∣
∣a 32 + 2 a (^12)
∣
∣
∣
} 2
If we puta 1 / 2 =0, we get the ratio as 3.10.27 The analysis is identical with that forπpreactions (problem 10.18). The
Σ’s are isospin triplet (T=1) with the third componentsT 3 =+ 1 , 0 ,− 1
forΣ+, Σ^0 , Σ−similar toπ+,π^0 ,π−. Further, theK+andK^0 form
a doublet (T= 1 /2) withT 3 =+ 1 /2 and− 1 /2, analogous topandn.
Therefore the result on the ratio of cross-section will be identical with that
for pion – nucleon reactions as in Problem 10.18
σ(π++p→Σ++K+):σ(π−+p→Σ−+K+):
σ(π−+p→Σ^0 +K^0 )
=∣
∣a 3 / 2∣
∣^2 :^1
9∣
∣a 3 / 2 + 2 a 1 / 2∣
∣^2 :^2
9∣
∣a 3 / 2 −a 1 / 2∣
∣^2
Ifa 1 / 2 <<a 3 / 2 , then the ratio becomes 9:1:2.10.28 Pions haveT=1 and nucleonsT=^1 / 2 , so that the resultant isospin both in
the initial state and the final state can beI=^1 / 2 or 3/2. Looking up the table
for Clebsch – Gordon Coefficients of 1× 1 /2 (Table 3.3) we can write
∣
∣π−p〉=√
1
3
∣
∣
∣
∣
3
2
,−
1
2
〉
−
√
2
3
∣
∣
∣
∣
1
2
,−
1
2
〉
∣
∣π^0 n〉=√
2
3
∣
∣
∣
∣
3
2
,−
1
2
〉
+
√
1
3
∣
∣
∣
∣
1
2
,−
1
2
〉
Therefore,〈
π−p|H|π−p〉
=^13 a 3 +^23 a 1〈
π^0 n∣
∣H
∣
∣π−p〉
=
√
2
3
a 3 −√
2
3
a 1